Chapter 6.5, Problem 21PPS

To determine

Whether JKLM is a rhombus, a rectangle or a square.

Answer to Problem 21PPS

QRST is neither a rectangle nor a rhombus.

Explanation of Solution

Given:

The vertices are:

  $J\left(-2,-1\right),K\left(-4,3\right),L\left(1,5\right),M\left(3,1\right).$

Calculation:

Plot the vertices on the graph:

According to Rectangle Theorem, if the diagonals of a parallelogram are congruent, then it is a rectangle.

According to Rhombi Theorem, if the diagonals of a parallelogram are perpendicular, then it is a rhombus.

According to Square Theorem, if the quadrilateral is both, a rectangle and a rhombus, then it is a square.

Consider the diagonals of JKLM:

Finding the length of JL:

  $\begin{array}{l}JL=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ JL=\sqrt{{\left(1-\left(-2\right)\right)}^2+{\left(5-\left(-1\right)\right)}^2}\\ JL=\sqrt{9+36}\\ JL=\sqrt{45}\end{array}$

Finding the length of KM:

  $\begin{array}{l}KM=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ KM=\sqrt{{\left(3-\left(-4\right)\right)}^2+{\left(1-3\right)}^2}\\ KM=\sqrt{49+4}\\ KM=\sqrt{53}\end{array}$

Both of these diagonals are not congruent, hence, QRST is not a rectangle.

Finding the slope of JL:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{5-\left(-1\right)}{1-\left(-2\right)}\\ m=\frac{6}{3}\\ m=2\end{array}$

Finding the slope of KM:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{1-3}{2-\left(-4\right)}\\ m=\frac{-2}{7}\\ m=-\frac{2}{7}\end{array}$

As, $2\left(-\frac{2}{7}\right)=-\frac{4}{7}$ , the diagonals are not perpendicular. It is not a rhombus.

Hence, QRST is neither a rectangle nor a rhombus.